Question

Answer by formula please ​

Answer 1

Explanation:

I honestly have no idea what you mean by answer by formula, but I'm going to give it my best. I began by squaring both sides to get:

(a² - b²) tan²θ = b² and then distributed to get:

a² tan²θ - b² tan²θ = b² and then got the b terms on the side to get:

a² tan²θ = b² + b² tan²θ and then changed the tans to sin/cos to get:

`(a^2sin^2\theta)/(cos^2\theta)=b^2+(b^2sin^2\theta)/(cos^2\theta)` and isolated the sin-squared on the left to get:

`a^2sin^2\theta=cos^2\theta(b^2+(b^2sin^2\theta)/(cos^2\theta))` and distributed to get:

***
`a^2sin^2\theta=b^2cos^2\theta+b^2sin^2\theta`*** and factored the right side to get:

`a^2sin^2\theta=b^2(sin^2\theta+cos^2\theta)` and utilized a trig Pythagorean identity to get:

`a^2sin^2\theta=b^2(1)` and then solved for sinθ in the following way:

`sin^2\theta=(b^2)/(a^2)` so

`sin\theta=(b)/(a)` This, along with the *** expression above will be important. I'm picking up at the *** to solve for cosθ:

`a^2sin^2\theta=b^2cos^2\theta+b^2sin^2\theta` and get the cos²θ alone on the right by subtracting to get:

`a^2sin^2\theta-b^2sin^2\theta=b^2cos^2\theta` and divide by b² to get:

`(a^2sin^2\theta)/(b^2)-sin^2\theta=cos^2\theta` and factor on the left to get:

`sin^2\theta((a^2)/(b^2)-1)=cos^2\theta` and take the square root of both sides to get:

`\sqrt{sin^2\theta((a^2)/(b^2)-1) }=cos\theta` and simplify to get:

`(sin\theta)/(b)√(a^2-b^2)=cos\theta` and go back to the identity we found for sinθ and sub it in to get:

`((b)/(a) )/(b)√(a^2-b^2)=cos\theta` and simplifying a bit gives us:

`(1)/(a)√(a^2-b^2)=cos\theta`

That's my spin on things....not sure if it's what you were looking for. If not.....YIKES